Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in CPP):

Solution:
#include <stdio.h>
#include <algorithm>
#include <string>
#include <vector>
using namespace std;

int N; double P[3030][3030];

int main()
{
	scanf ("%d",&N);
	P[0][0] = 1;
	for (int i=0;i<N;i++){
		double p; scanf ("%lf",&p);
		for (int j=0;j<=i;j++){
			P[i+1][j+1] += P[i][j] * p;
			P[i+1][j] += P[i][j] * (1 - p);
		}
	}

	double ans = 0;
	for (int i=N,j=0;i>j;i--,j++) ans += P[N][i];
	printf ("%.12lf\n",ans);

	return 0;
}